We can prove that if $h(s) \neq 0$ then $f(s)=0$, $f(b-s)=0$ (I know that if $f(s)=0$ then $f(b-s)=0$). However the case $h(s)=0$ does not imply this case directely. But we can see also that we can have $f(s)=0$, $f(b-s)=0$ if $h(s)=0$.

(a) My question is how to deal with the case when $h(s)=0$.

(b) How I can solve this functional equation: $g(1-s)g(s)=1$ with respect to $g$ for all $s$.

If $g(s)>0$ and $h(s)\in(-\pi,\pi]$, shouldn't the codomain of $g$ and $h$ be $\mathbb R$? Also, is the equation supposed to be true for all $s\in\mathbb C$ and all $b\in\mathbb R$, or for all $s$ and a particular $b$, or what?
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RahulDec 30 '12 at 13:06

Yes, Done. The equation hlods for all $s$ and some $b$ including $b=1$. I am searching for the roots of $f(s)=0$ when $b=1$.
–
ZE1Dec 30 '12 at 13:14

1 Answer
1

Fix $b \in \mathbb{R}$ and suppose that $f(s) = g(s)\exp(ih(s))f(b-s)$ for all $s \in \mathbb{C}$. As $g(s) > 0$ and $\exp(ih(s)) \neq 0$, $f(s) = 0$ if and only if $f(b - s) = 0$. Depending on $f$, there may not be any zeroes, but if there is a zero $s_0$, then $b - s_0$ is another zero.